# Consider a non-uniform rod with length L on the x-axis between x = 0 and x = L. The linear mass...

## Question:

Consider a non-uniform rod with length {eq}\, L \, {/eq} on the x-axis between {eq}\, x \, = \, 0 \, {/eq} and {eq}\, x \, = \, L {/eq}. The linear mass density of the rod varies with x and is given by the function {eq}\, \lambda(x) \, = \, \lambda_0 \, \left (1 \, + \, \displaystyle \frac{x^{2}}{L^{2}} \right ) {/eq}, where {eq}\, \lambda_0 \, {/eq} is a positive constant. In terms of {eq}\, \lambda_0 \, {/eq} and {eq}\, L {/eq}, what are (a) the mass of the rod and (b) the location of the center-of-mass of the rod.

## Center of mass:

The center of mass of the system of point masses is the average position of all masses weighted according to their masses. For the system of **N** point masses, the center of mass is given by the following equation: {eq}\vec r_{cm} = \dfrac {\sum \limits_{i=1}^N m_i \vec r_i}{\sum \limits_{i=1}^N m_i} {/eq}. In the case of continuous mass distribution, the summation shall be replaced by the integration.

## Answer and Explanation:

a) The total mass of the rod is given by the equation:

{eq}m = \displaystyle \int \limits_0^L \lambda(x) dx {/eq}

Computing the integral, we...

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from High School Physics: Help and Review

Chapter 1 / Lesson 10